Remove object from array based on array of some property of that object2019 Community Moderator...
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Remove object from array based on array of some property of that object
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Remove object from array based on array of some property of that object
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I have an array of objects (objList) that each has "id" property.
I have an array of strings (idsToRemove), representing IDs of the objects to remove from objList.
I find some solution but I fear it's slow, especially with the large list of objects with lots of properties.
Is there more efficient way to do this?
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
for (var i = 0, len = idsToRemove.length; i < len; i++) {
objList = objList.filter(o => o.id != idsToRemove[i]);
}
console.log(objList);javascript arrays performance
edited 6 hours ago
Uwe Keim
27.6k32132213
asked 8 hours ago
DaliborDalibor
482318
add a comment |
I have an array of objects (objList) that each has "id" property.
I have an array of strings (idsToRemove), representing IDs of the objects to remove from objList.
I find some solution but I fear it's slow, especially with the large list of objects with lots of properties.
Is there more efficient way to do this?
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
for (var i = 0, len = idsToRemove.length; i < len; i++) {
objList = objList.filter(o => o.id != idsToRemove[i]);
}
console.log(objList);javascript arrays performance
edited 6 hours ago
Uwe Keim
27.6k32132213
asked 8 hours ago
DaliborDalibor
482318
add a comment |
I have an array of objects (objList) that each has "id" property.
I have an array of strings (idsToRemove), representing IDs of the objects to remove from objList.
I find some solution but I fear it's slow, especially with the large list of objects with lots of properties.
Is there more efficient way to do this?
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
for (var i = 0, len = idsToRemove.length; i < len; i++) {
objList = objList.filter(o => o.id != idsToRemove[i]);
}
console.log(objList);javascript arrays performance
edited 6 hours ago
Uwe Keim
27.6k32132213
asked 8 hours ago
DaliborDalibor
482318
I have an array of objects (objList) that each has "id" property.
I have an array of strings (idsToRemove), representing IDs of the objects to remove from objList.
I find some solution but I fear it's slow, especially with the large list of objects with lots of properties.
Is there more efficient way to do this?
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
for (var i = 0, len = idsToRemove.length; i < len; i++) {
objList = objList.filter(o => o.id != idsToRemove[i]);
}
console.log(objList);var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
for (var i = 0, len = idsToRemove.length; i < len; i++) {
objList = objList.filter(o => o.id != idsToRemove[i]);
}
console.log(objList);var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
for (var i = 0, len = idsToRemove.length; i < len; i++) {
objList = objList.filter(o => o.id != idsToRemove[i]);
}
console.log(objList);javascript arrays performance
javascript arrays performance
edited 6 hours ago
Uwe Keim
27.6k32132213
asked 8 hours ago
DaliborDalibor
482318
edited 6 hours ago
Uwe Keim
27.6k32132213
asked 8 hours ago
DaliborDalibor
482318
edited 6 hours ago
Uwe Keim
27.6k32132213
edited 6 hours ago
Uwe Keim
27.6k32132213
edited 6 hours ago
Uwe Keim
27.6k32132213
27.6k32132213
asked 8 hours ago
DaliborDalibor
482318
asked 8 hours ago
DaliborDalibor
482318
asked 8 hours ago
DaliborDalibor
482318
482318
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Turn the idsToRemove into a Set so that you can use Set.prototype.has (an O(1) operation), and .filter the objList just once, so that the overall complexity is O(n) (and you only iterate over the possibly-huge objList once):
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);Note that Array.prototype.includes and Array.prototype.indexOf operations are O(N), not O(1), so if you use them instead of a Set, they may take significantly longer.
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
add a comment |
You can use Array.includes which check if the given string exists in the given array and combine it with an Array.filter.
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
add a comment |
Simply use Array.fliter()
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
add a comment |
You don't need two nested iterators if you use a built-in lookup function
objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);
Documentation:
Array.prototype.indexOf()
Array.prototype.includes()
answered 8 hours ago
Charlie HCharlie H
9,48442653
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Turn the idsToRemove into a Set so that you can use Set.prototype.has (an O(1) operation), and .filter the objList just once, so that the overall complexity is O(n) (and you only iterate over the possibly-huge objList once):
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);Note that Array.prototype.includes and Array.prototype.indexOf operations are O(N), not O(1), so if you use them instead of a Set, they may take significantly longer.
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
add a comment |
Turn the idsToRemove into a Set so that you can use Set.prototype.has (an O(1) operation), and .filter the objList just once, so that the overall complexity is O(n) (and you only iterate over the possibly-huge objList once):
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);Note that Array.prototype.includes and Array.prototype.indexOf operations are O(N), not O(1), so if you use them instead of a Set, they may take significantly longer.
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
add a comment |
Turn the idsToRemove into a Set so that you can use Set.prototype.has (an O(1) operation), and .filter the objList just once, so that the overall complexity is O(n) (and you only iterate over the possibly-huge objList once):
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);Note that Array.prototype.includes and Array.prototype.indexOf operations are O(N), not O(1), so if you use them instead of a Set, they may take significantly longer.
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
Turn the idsToRemove into a Set so that you can use Set.prototype.has (an O(1) operation), and .filter the objList just once, so that the overall complexity is O(n) (and you only iterate over the possibly-huge objList once):
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);Note that Array.prototype.includes and Array.prototype.indexOf operations are O(N), not O(1), so if you use them instead of a Set, they may take significantly longer.
var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);var idsToRemove = ["3", "1"];
var objList = [{
id: "1",
name: "aaa"
},
{
id: "2",
name: "bbb"
},
{
id: "3",
name: "ccc"
}
];
const set = new Set(idsToRemove);
const filtered = objList.filter(({ id }) => !set.has(id));
console.log(filtered);answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
edited 8 hours ago
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
answered 8 hours ago
CertainPerformanceCertainPerformance
91.5k165280
91.5k165280
add a comment |
add a comment |
You can use Array.includes which check if the given string exists in the given array and combine it with an Array.filter.
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
add a comment |
You can use Array.includes which check if the given string exists in the given array and combine it with an Array.filter.
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
add a comment |
You can use Array.includes which check if the given string exists in the given array and combine it with an Array.filter.
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
You can use Array.includes which check if the given string exists in the given array and combine it with an Array.filter.
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
},
];
const filteredObjList = objList.filter(x => !idsToRemove.includes(x.id));
console.log(filteredObjList);answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
answered 8 hours ago
Grégory NEUTGrégory NEUT
9,38621940
9,38621940
add a comment |
add a comment |
Simply use Array.fliter()
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
add a comment |
Simply use Array.fliter()
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
add a comment |
Simply use Array.fliter()
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
Simply use Array.fliter()
const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);const idsToRemove = ['3', '1'];
const objList = [{
id: '1',
name: 'aaa',
},
{
id: '2',
name: 'bbb',
},
{
id: '3',
name: 'ccc',
}
];
const res = objList.filter(value => !idsToRemove.includes(value.id));
console.log("result",res);answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
answered 8 hours ago
Khyati SharmaKhyati Sharma
1306
1306
add a comment |
add a comment |
You don't need two nested iterators if you use a built-in lookup function
objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);
Documentation:
Array.prototype.indexOf()
Array.prototype.includes()
answered 8 hours ago
Charlie HCharlie H
9,48442653
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
add a comment |
You don't need two nested iterators if you use a built-in lookup function
objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);
Documentation:
Array.prototype.indexOf()
Array.prototype.includes()
answered 8 hours ago
Charlie HCharlie H
9,48442653
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
add a comment |
You don't need two nested iterators if you use a built-in lookup function
objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);
Documentation:
Array.prototype.indexOf()
Array.prototype.includes()
answered 8 hours ago
Charlie HCharlie H
9,48442653
You don't need two nested iterators if you use a built-in lookup function
objList = objList.filter(o => idsToRemove.indexOf(o.id) < 0);
Documentation:
Array.prototype.indexOf()
Array.prototype.includes()
answered 8 hours ago
Charlie HCharlie H
9,48442653
edited 8 hours ago
answered 8 hours ago
Charlie HCharlie H
9,48442653
answered 8 hours ago
Charlie HCharlie H
9,48442653
answered 8 hours ago
Charlie HCharlie H
9,48442653
9,48442653
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
add a comment |
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
This leaves those I wanted to delete in the list, it should be indexOf == -1, isn't it?
– Dalibor
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
Yes. Answer edited
– Charlie H
8 hours ago
add a comment |
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